finding the molecular formula

Question

An unknown compound has a formula of CxHyOz. You burn 0.1523 g of the compound and isolated 0.3718 g of CO2, and 0.1522 g of H2O. what is the empirical formula of the compound? If the molar mass is 72.1g/mol, what is the molecular formula?

Answer 1

The way to address this problem is to calculate the amount of C and H in the products.   This will tell you the amount of C and H in the unknown compound.   The remaining mass of the compound will be O.

 

.3718g of CO2 * 1 mole/44g =  0.00845 moles of CO2 so the initial product included  0.00845 moles of C 

0.1522  g H2) * 1gm/mole H2O * 2mole H/mole H2)  gives 0.0169 moles of H

 

convert the carbon and H to grams and subtract to find the amount of oxygetn

 

1* 0.0169 + 12*0.00845 = .1183 grams.   .1523-.1183 = 0.034 gram of O2 = 0.00215 moles of O2.

 

divide each of the number of moles by the smallest number 0.00215 gives

 

O  1.00

H   7.86

C   3.93 

 

Rounding gives an empirical formula   C₄H₈O.    The corresponding empircal weight is 72.0.  Since this is also the known molecular weight, the empirical formula is also the molecular formula.